Question: In the diagram, $PQ$ and $RS$ are diameters of a circle with radius 4.  If $PQ$ and $RS$ are perpendicular, what is the area of the shaded region?

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size(120);
import graph;
fill((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,mediumgray);
fill(Arc((0,0),sqrt(2),45,135)--cycle,mediumgray);fill(Arc((0,0),sqrt(2),225,315)--cycle,mediumgray);
draw(Circle((0,0),sqrt(2)));
draw((-1,-1)--(1,1)--(1,-1)--(-1,1)--cycle);
label("$P$",(-1,1),NW); label("$R$",(1,1),NE); label("$S$",(-1,-1),SW); label("$Q$",(1,-1),SE);

[/asy]
Explanation: Diameters $PQ$ and $RS$ cross at the center of the circle, which we call $O$.

The area of the shaded region is the sum of the areas of $\triangle POS$ and $\triangle ROQ$ plus the sum of the areas of sectors $POR$ and $SOQ$.

Each of $\triangle POS$ and $\triangle ROQ$ is right-angled and has its two perpendicular sides of length 4 (the radius of the circle).

Therefore, the area of each of these triangles is $\frac{1}{2}(4)(4)=8$.

Each of sector $POR$ and sector $SOQ$ has area $\frac{1}{4}$ of the total area of the circle, as each has central angle $90^\circ$ (that is, $\angle POR = \angle SOQ = 90^\circ$) and $90^\circ$ is one-quarter of the total central angle.

Therefore, each sector has area $\frac{1}{4}(\pi(4^2))=\frac{1}{4}(16\pi)=4\pi$.

Thus, the total shaded area is $2(8)+2(4\pi)=\boxed{16+8\pi}$.